Answer
(a)
Increasing on $(-\sqrt{2}, \sqrt{2})$
Decreasing on $(-\infty, -\sqrt{2})\cup(\sqrt{2}, \infty)$
(b)
No absolute maximum.
Local maximum at $(\sqrt{2},4\sqrt{2})$.
No absolute minimum.
Local minimum at $(-\sqrt{2},-4\sqrt{2})$.
Work Step by Step
$f$ is defined everywhere.
$ f'(\theta)=6-3\theta^{2}=3(2-\theta^{2}),\quad$ defined everywhere.
$f'(\theta)=0$ for $\theta=\pm\sqrt{2}$
Critical points: $\theta=\pm\sqrt{2}$
$f(-\sqrt{2})=-4\sqrt{2},\qquad f(\sqrt{2})=4\sqrt{2}.$
The end behavior of a polynomial is dictated by the leading term.
Using testpoints in the intervals between the critical points,
$f'(-2) \lt 0$
$f'(0) \gt 0$
$f'(2) \lt 0$
$ \begin{array}{l}
f':\\
\\
\\
f:\\
\end{array} \quad \begin{array}{ccccccccccc}
-\infty& &-\sqrt{2} & &\sqrt{2} & &\infty \\
{(} &-- &| &++ &| & -- &) \\ \hline
(+\infty)&\searrow & & \nearrow & 4\sqrt{2} & \searrow & \\
& & -4\sqrt{2} & & & & (-\infty) \end{array}$
(a)
Increasing on $(-\sqrt{2}, \sqrt{2})$
Decreasing on $(-\infty, -\sqrt{2})\cup(\sqrt{2}, \infty)$
(b)
No absolute maximum.
Local maximum at $(\sqrt{2},4\sqrt{2})$.
No absolute minimum.
Local minimum at $(-\sqrt{2},-4\sqrt{2})$.
